What’s the difference between a formula and an equation? Take the examples,

\(V = 11.3 V \hspace{ 10pt } and \hspace{ 10pt } V = RI.\)

The first is a formula. It requires the units \([V]\) or [volts] to be stated, because the statement "\(V = 12\)" could mean \(12 mV\) or \(12 kV\) or \(12 p.u.\), or even \(12 V\)! The essential function of a formula is data or information, rather than understanding. It conveys a fact without any explanation. We often use formulas in work processes without thinking too much about their origins.

The second is an equation in the sense that it equates the left-hand side with the right-hand side. It conveys an understanding of voltage and current in a circuit containing resistance \(R\) : Ohm’s law. If the parameters are expressed in a consistent set of units, there is no need to include the units in the equation. If resistance is in \([ohm]\) while current is in \([A]\), then voltage \(RI\) will be in \([V]\). An equation works both ways : we can write \(RI = V\). But it is almost meaningless to write \(11.3 = V\).

A mathematician might say that an equation never requires units because it expresses an immutable truth. It is hard to quarrel with that argument in the case of an equation such as

\(e^{jx} = \cos x + j \sin x\)

where \(j^2+1 = 0\). Whoever would complain that de Moivre forgot the units? An even more basic equation from geometry is

\(r^2 = x^2 + y^2\)

expressing the relationship between the hypotenuse \(r\), the base \(x\) and the perpendicular \(y\) in a right-angled triangle. Did Pythagoras forget his units as well? We could write this equation as

\(r^2 = 645 \cdot 16x^2 + y^2.\)

In this case \(x\) is measured in \([in]\) (inches) while \(r\) and \(y\) are measured in \([mm]\). We now have a formula : quite a horrible one, because of the mixed units. Is it still an equation? Whereas the form \(r^2 = x^2 + y^2\) is tacitly assumed to be always true, and is always true in a consistent set of units, the second form is true only if \(x, y\) and \(r\) are in the particular units stated.

An example from electromagnetics is

\(\mu = B/H\)

which defines the permeability \(\mu\) of a homogeneous anisotropic linear material in terms of the magnetic flux-density \(B\) and the magnetic field-strength \(H\). In a vacuum (or very nearly, in air), \(\mu\) has the value \(4\pi × 10^{-7} [H/m]\) which we call \(\mu_0\). But this is meaningful only in the Système Internationale, in which \(B\) is in \([T]\) and \(H\) is in \([A/m]\). How we get \([H/m]\) from \([T]/[A/m]\) is by no means obvious : it requires an understanding of the electromagnetics. Physicists did a lot of hard work to make \(\mu_0 = 4\pi × 10^{-7} [H/m]\). Thank you! In so-called c.g.s. units \(B\) is in \([G]\) or [gauss] while \(H\) is in [Oe] or [oersted], and in air \(\mu = 1\), a very different value. If \(B\) is in kilo-lines/in² \([kl/in^2]\) and \(H\) is in \([A/in]\), \(\mu = 3 \cdot 19\) “units”. All these units are still in use today, somewhere. It is clear that engineers using such formulas and equations must be careful to state the units of all parameters, and never assume that the reader will know the units.

Many familiar formulas in engineering are often linearized approximations which omit many details and complications. For example, permeability \(\mu\) can be defined as \(B/H\) or \(dB/dH\), that is, either the total or the incremental permeability. Both definitions are important and both have practical implications. In an anisotropic material we may have to write \(\mu\) as a matrix (sometimes identified as a tensor) relating three vector components of \(B\) to three vector components of \(H\).

A common formula used in the design of transformers or AC motors is the no-load voltage

\(E = 4\cdot 44 k_w \Phi Nf \hspace{ 5pt } [V rms]\)

This classic engineering formula expresses one of the most fundamental truths of AC power engineering — Faraday’s law; but it would take at least a paragraph and maybe a page to explain the units and meaning of every parameter. Where does that \(4 \cdot 44\) coefficient come from? Not from Faraday himself, surely. And under what conditions is the formula exactly true? One could say that as a practical formula it is never exactly true : it may be approximately true within the limits of experimental error and other uncertainties.

Another detail in equation-writing is the order of the parameters. This is not simply a matter of style; particularly with matrix equations it can make the difference between right and wrong, and a wrongly-written formula with matrices or complex numbers may give an incorrect result, or even be impossible to calculate. That’s why in the second equation \(RI\) is written instead of \(IR\). \(R\) is viewed as an operator which “operates” on the current to produce the voltage. In an inductor or reactor, we would have \(ZI = (R + jX) I\) where \(I\) is complex (a phasor), and \(Z\) is the AC impedance. It is almost a matter of “equation etiquette” to be fussy about the order of parameters, even when \(IR = RI\).

What about the expression \(I^2R\)? Is that bad etiquette? Does anyone ever write \(RI^2\)? In this case we could use the “operator” idea and write \(IRI\). The resistance operates on the current to produce a voltage \(RI\), and then we drive a current \(I\) (the same current) through that voltage to produce the power loss \(IV\), which is \(IRI\). Pedantic? Maybe. Unconventional? Certainly. But suppose that “the current” is a column vector \(i\) of several currents in a nework, in which \(R\) is a square matrix of resistances (including self-resistances on the diagonal and mutual resistances off the diagonal). Then the power loss can be written and calculated by \(i’Ri\), where \(i’\) is the transpose of \(i\). The result is a scalar, the total power loss. So if we write \(IRI\) for a network with only one element, and \(R\) is a \(1 × 1\) matrix if we want to be pedantic, we are tipping our caps to the more general case of an extended network. From this point of view, which is excessively academic in most instances, \(I^2R\) is colloquial, vernacular, — “shop talk”. But OK.

Just one last point. In an AC network where the currents and voltages are phasors, the expression becomes \(I’^*RI\), (where * means the complex conjugate). Which of course we call the \(I^2R\) losses! Don’t forget to use the RMS values of current!